Answer:
0.51
Step-by-step explanation:
Subtract the post from the pre:
[tex]\left[\begin{array}{cccccc}Student&1&2&3&4&5\\Pre-test&11&9&10&14&10\\Post-test&18&17&19&20&18\\Pre-Post&-7&-8&-9&-6&-8\end{array}\right][/tex]
The mean of the differences is:
μ = (∑x)/n
μ = (-7−8−9−6−8)/5 = -7.6.
The standard deviation is:
s = √([∑(x−μ)] / (n−1))
s = √([(-7+7.6)²+(-8+7.6)²+(-9+7.6)²+(-6+7.6)²+(-8+7.6)²] / (5−1))
s = 1.14
The standard error is:
SE = s / √n
SE = 1.14 / √5
SE = 0.51
